# Convergens of a powerseries

1. May 14, 2006

### Mathman23

Hi

I'm suppose to show that the powerseries

$$\sum_{n=0} ^{\infty} \frac{2^n}{2n+1} z^{2n+1}$$

converge for all z \in mathbb{C}

By using the Ratio test:

I get

$$\sum_{n=0} ^{\infty} \frac{2^n}{2n+1} z^{2n+1} = \frac{2(2n+1)}{2n+3} |z^2| = \frac{2}{3}|z^2|, n \rightarrow \infty$$

I guess that can be re-written as |z^2| < 3/2 ???

But does that mean the ratio test fails?

Sincerley

Fred

Last edited: May 14, 2006
2. May 14, 2006

### Curious3141

Firstly, are you sure that series converges for *all* z in C with no restrictions on the magnitude?

Secondly, when doing the ratio test, you should write the sequence as being equal to the ratio of two successive terms (I mean the notation is incorrect - you're effectively equating the series sum to the ratio when you put an equal sign there).

3. May 14, 2006

### Jameson

I'm going to do the Ratio Test as I type to check your work.

Calling $$b_n=\frac{a_{n+1}}{a_n}$$ I get that $$b_n=\frac{(2^{n+1})(z^{2n+3})}{2n+3} \times \frac{2n+1}{(2^n)(z^{2n+1})}$$.

Now for simplification.

$$b_n=\frac{2*2^n*z^{2n}*z^3}{2n+3} \times \frac{2n+1}{2^n*z^{2n}*z}$$.

$$b_n=\frac{(4n+2)*z^2}{2n+3}$$. Now take the limit as $$n \to \infty$$.

Ok so you've done your work right until the applying the limit. I get that the limit as n approaches infinity is $$2z^2$$. Now you must find z such that $$|2z^2|<1$$

Last edited: May 14, 2006
4. May 15, 2006

### Mathman23

Dear Jameson,

Thank You for checking and correcting my work :)

Is preferable to

$$|z^2|<1/2$$ ?

Sincerely Yours
Fred

5. May 15, 2006

### HallsofIvy

Staff Emeritus
First do not write
$$\sum_{n=0} ^{\infty} \frac{2^n}{2n+1} z^{2n+1} = \frac{2(2n+1)}{2n+3} |z^2|$$
The infinite sum is not equal to the ratio!!

Second, are you asked to show that the series converges for all z or to find the radius of convergence?

I don't know what you mean by "Is preferable to $|z|^2< \frac{1}{2}$?". As Jameson showed, the correct ratio is $|2z^2|< 1$. Since 2 is positive, that is the same as $2|z^2|< 1$ or
$|z|^2< \frac{1}{2}$. Finally, since |z| is a positive real number, $|z|< \frac{1}{\sqrt{2}}$.

No, the ratio test does not "fail" but it does show that this series does not converge for all z.

Last edited: May 15, 2006
6. May 15, 2006

### Mathman23

Hello Hall,

I suppose to show that the series converge for all z. Since using ratio test shows that the series converge for $$|z| < \frac{1}{\sqrt(2)}$$, then I need to use the comparison test?

Sincerely Yours

Fred

7. May 15, 2006

### HallsofIvy

Staff Emeritus
No! What the ratio test shows is that this series does not converge for all z! It has radius of convergence $\frac{1}{\sqrt{2}}$.

If you are not convinced, take z= 1. Does
$$\sum_{n=0} ^{\infty} \frac{2^n}{2n+1}$$
converge?

Does the sequence
$$\frac{2^n}{2n+1}$$
even converge to 0?

Last edited: May 15, 2006
8. May 15, 2006

### Mathman23

Thanks now I get ;)

Yours
Fred

9. May 17, 2006

### Mathman23

Hello I discovered that I made an error in the post,

The original question should have looked like this:

Given a powers series

$$\sum_{n=0} ^\infty \frac{2^n}{(2n+1)!!} z^{2n+1}$$

Show that the series converge for all $$z \in \mathbb{C}$$

By the ratio test,

I get that

$$\mathop {\lim }\limits_{n \to \infty } \left| \frac{4(n+1) \cdot n! \cdot z^{2n+3}}{(2n+3)!} \right| =\mathop {\lim }\limits_{n \to \infty } \left| \frac{4(n+1) \cdot n!}{(2n+3)!} \right| \left| z^{3} \right|$$

Since $$n \rightarrow \infty$$ then

$$\mathop {\lim }\limits_{n \to \infty } \left| \frac{4(n+1) \cdot n! \cdot z^{2n+3}}{(2n+3)!} \right| \rightarrow 0.$$

Therefore it converges for all z?

Sincerely
Fred

Last edited: May 17, 2006
10. May 17, 2006

### Curious3141

I'm sorry, but I really cannot understand the algebra you're doing.

How did the '4', the '(n+1)!' etc. come into it? How did you get a z^3 term in the ratio?

First of all, please note that $$(2n+1)!! \neq (2n+1)!$$

For a definition of the double factorial, please see your earlier thread entitled Powersums and ODEs and my exchange with Mathman. You're dealing with the same series as in that thread, and that is a double factorial.

What I get is like this. $$T_k$$ represents the kth term in the series.

$$r = \frac{T_{n+1}}{T_n} = \frac{2z^2}{2n+3}$$

$$lim_{n \rightarrow \infty} r = lim_{n \rightarrow \infty} \frac{2z^2}{2n} = lim_{n \rightarrow \infty} \frac{z^2}{n} = 0$$

so the series converges by the ratio test.

Last edited: May 17, 2006